Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int MAX(int a, int b) { return (a > b) ? a : b; }
int MIN(int a, int b) { return (a < b) ? a : b; }
int ABS(int x) { return (x > 0) ? x : -x; }
int left(int x) { return (x << 1); }
int right(int x) { return (x << 1) + 1; }
int lsone(int x) { return (x & (-x)); }
int n, q, al, ar;
double s, a, b, segl[300000], segr[300000], sumSeg[300000], x, y;
int main() {
  scanf("%lf %lf %lf", &s, &a, &b);
  scanf("%d", &n);
  for (long long int i = 0; i < (long long int)n; i++)
    scanf("%lf %lf", &segl[i], &segr[i]);
  for (long long int i = 0; i < (long long int)n; i++) {
    sumSeg[i] = segr[i] - segl[i];
    if (i) sumSeg[i] += sumSeg[i - 1];
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%lf %lf", &x, &y);
    double ap, bp, lenp = 0.00, ans = 0.00;
    ap = a + (-s) * ((x - a) / (y - s));
    bp = b + (-s) * ((x - b) / (y - s));
    int l = 0, r = n - 1, al, ar;
    while (l < r) {
      int mid = (l + r) / 2;
      if (segr[mid] < ap)
        l = mid + 1;
      else
        r = mid;
    }
    al = l;
    l = 0;
    r = n - 1;
    while (l < r) {
      int mid = (l + r) / 2 + 1;
      if (segl[mid] <= bp)
        l = mid;
      else
        r = mid - 1;
    }
    if (segl[r] > bp) r--;
    ar = r;
    if (ar >= 0) ans += sumSeg[ar];
    if (al > 0) ans -= sumSeg[al - 1];
    if (al >= 0 && ap > segl[al]) {
      ans -= ap - segl[al];
    }
    if (ar >= 0 && bp < segr[ar]) {
      ans -= segr[ar] - bp;
    }
    ans *= (y - s) / (y);
    if (ans < 0.00) ans = 0.00;
    printf("%.15lf\n", ans);
  }
}